TSQL仅选择最高的credentail_Sql_Sql Server_Tsql

TSQL仅选择最高的credentail

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TSQL仅选择最高的credentail,sql,sql-server,tsql,Sql,Sql Server,Tsql,我有一个查询返回单个服务的多行，因为一个人可能有多个凭据。在医学领域，你会保留一些证书，但为了简单起见，我只会使用标准证书，如博士、硕士、硕士、学士、学士我需要知道忽略Z_ServiceLedger.clientvisit_id在层次结构中具有任何凭据的行的最简单方法。因此，如果一名员工提供服务，他拥有博士学位和硕士学位，则只返回博士学位的行，如果他拥有博士学位，则只返回博士学位的行。我们有大约50个凭证，所以如果我使用每个凭证的用例，您可以看到这会变得多么混乱，我希望有更好的方法来避免这种情

我有一个查询返回单个服务的多行，因为一个人可能有多个凭据。在医学领域，你会保留一些证书，但为了简单起见，我只会使用标准证书，如博士、硕士、硕士、学士、学士

我需要知道忽略Z_ServiceLedger.clientvisit_id在层次结构中具有任何凭据的行的最简单方法。因此，如果一名员工提供服务，他拥有博士学位和硕士学位，则只返回博士学位的行，如果他拥有博士学位，则只返回博士学位的行。我们有大约50个凭证，所以如果我使用每个凭证的用例，您可以看到这会变得多么混乱，我希望有更好的方法来避免这种情况

以下是我当前的查询：

SELECT DISTINCT
    SUM(CASE WHEN v.non_billable = 0 THEN v.duration ELSE 0 END) / 60 AS billable_hours,
    SUM(CASE WHEN (v.non_billable = 0 AND Z_ServiceLedger.payer_id = 63) THEN v.duration ELSE 0 END) / 60 AS billable_mro_hours,
    Credentials.credentials
FROM 
    Z_ServiceLedger
INNER JOIN 
    ClientVisit v ON Z_ServiceLedger.clientvisit_id = v.clientvisit_id
LEFT JOIN 
    Employees ON v.emp_id = Employees.emp_id
LEFT JOIN 
    EmployeeCredential ON Employees.emp_id = EmployeeCredential.emp_id
LEFT JOIN 
    Credentials ON Credentials.credential_id = EmployeeCredential.credential_id
WHERE 
    v.rev_timein <= CASE
                       WHEN EmployeeCredential.end_date IS NOT NULL 
                          THEN EmployeeCredential.end_date 
                          ELSE GETDATE()
                       END 
    AND v.rev_timein >= @param1 
    AND v.rev_timein < DateAdd(d, 1, @param2) 
    AND Z_ServiceLedger.amount > 0 
    AND v.splitprimary_clientvisit_id IS NULL
    AND v.gcode_primary_clientvisit_id IS NULL 
    AND v.non_billable = 0
    AND v.non_billable = 'FALSE' 
    AND v.duration / 60 > 0 
    AND Z_ServiceLedger.action_type NOT IN ('SERVICE RATE CHANGE', 'CLIENT STATEMENT') 
    AND (EmployeeCredential.is_primary IS NULL OR EmployeeCredential.is_primary != 'False') 
    AND v.client_id != '331771 '
GROUP BY 
    Credentials.credentials,
    v.non_billable
ORDER BY 
    Credentials.credentials

选择DISTINCT
总和（如果v.non_bilable=0，则v.duration否则0 END）/60作为计费小时，
总和（当（v.non_bilable=0和Z_serviceloger.payer_id=63）然后v.duration ELSE 0 END）/60作为可计费的mro_小时，
凭证
从…起
Z_服务分类账
内连接
Z_ServiceLedger上的ClientVisit v.ClientVisit_id=v.ClientVisit_id
左连接
v.emp_id上的员工=Employees.emp_id
左连接
EmployeeCredential ON Employees.emp\u id=EmployeeCredential.emp\u id
左连接
凭据上的凭据。凭据\u id=员工凭据。凭据\u id
哪里
v、 rev_timein=@param1
和v.rev_timein0
并且v.splitprimary\u clientvisit\u id为空
并且v.gcode_primary_clientvisit_id为空
和v.non_bilable=0
和v.non_bilable='FALSE'
和v.duration/60>0
和Z_ServiceLedger.action_类型不在（'SERVICE RATE CHANGE'，'CLIENT STATEMENT'）
和（EmployeeCredential.is_primary为NULL或EmployeeCredential.is_primary！=“False”）
和v.client_id！='331771 '
分组
凭证，凭证，
v、 不计费
订购人
凭证

一些别名和格式确实揭示了一些主要的逻辑缺陷。在where子句中至少有两个谓词在逻辑上将左连接转换为内部连接。这完全是瞎猜，因为从你们今天的两个问题来看，我们没有任何实际的表格或样本数据

不过，最大的问题是where子句试图获取行v.non_billable=0，其中它等于'FALSE'。不可能两者都有

Select sum(Case When v.non_billable = 0 Then v.duration Else 0 End) / 60 As billable_hours
    , sum(Case When (v.non_billable = 0 And sl.payer_id = 63) Then v.duration Else 0 End) / 60 As billable_mro_hours
    , c.credentials
From Z_ServiceLedger sl
Inner Join ClientVisit v On sl.clientvisit_id = v.clientvisit_id
Left Join Employees e On v.emp_id = e.emp_id
Left Join EmployeeCredential ec On e.emp_id = ec.emp_id
                        --if you leave these predicates in the where clause you have turned your left join into an inner join.
                        AND v.rev_timein <= isnull(ec.end_date, GetDate()) 
                        and (ec.is_primary Is Null Or ec.is_primary != 'False')                     
Left Join Credentials c On c.credential_id = ec.credential_id
Where v.rev_timein >= @param1 
    And v.rev_timein < DateAdd(day, 1, @param2) 
    And v.splitprimary_clientvisit_id Is Null
    And v.gcode_primary_clientvisit_id Is Null
    --you need to pick one value for v.non_billable. It can't be both 0 and 'FALSE' at the same time. 
    And v.non_billable = 0 
    And v.non_billable = 'FALSE' 
    --And v.duration / 60 > 0 
    and v.duration > 60 --this is the same thing and is SARGable
    And sl.amount > 0 
    And sl.action_type NOT IN ('SERVICE RATE CHANGE', 'CLIENT STATEMENT') 
    And v.client_id != '331771 '
Group By c.credentials
    , v.non_billable
Order By c.credentials

选择sum（如果v.non_bilable=0，则v.duration Else 0 End）/60作为计费小时数
，总和（当（v.non_bilable=0和sl.payer_id=63）时，则v.duration Else 0 End）/60作为可计费的mro_小时数
，c
来自Z_ServiceLedger sl
sl.ClientVisit\u id=v.ClientVisit\u id上的内部联接客户端访问v
左键在v.emp_id=e.emp_id上加入员工e
左加入e.emp_id=ec.emp_id上的EmployeeCredential ec
--如果将这些谓词保留在where子句中，则已将左连接变成内部连接。
和v.rev_timein=@param1
和v.rev_timein0
v.duration>60——这是同样的事情，是可以接受的
和sl.amount>0
和sl.action_类型不在（'服务费率变化'，'客户声明'）
和v.client_id！='331771 '
按c.凭据分组
，v.不收费
按c.证书订购

编辑：修改查询以添加CTE，使用

FROM（VALUES（…）

表值构造函数语法计算

凭证等级。这适用于SQL 2008+。（）

首先，我将构建一段非常简单的数据
设置：
CREATE TABLE Employees ( emp_id int, emp_name varchar(20) ) ;
INSERT INTO Employees (emp_id, emp_name)
VALUES (1,'Jay'),(2,'Bob')
;

CREATE TABLE Credentials ( credential_id int, credentials varchar(20), credential_rank int ) ;
INSERT INTO Credentials (credential_id, credentials, credential_rank)
VALUES (1,'BA',3),(2,'MA',2),(3,'PhD',1)
;

CREATE TABLE EmployeeCredential (emp_id int, credential_id int, is_primary bit, end_date date )
INSERT INTO EmployeeCredential (emp_id, credential_id, is_primary, end_date)
VALUES 
    ( 1,2,null,'20200101' )
  , ( 1,3,0,'20200101' ) /* NON-PRIMARY */
  , ( 1,1,1,'20100101' ) /* EXPIRED CRED */
  , ( 2,3,null,'20200101' )
  , ( 2,3,1,'20200101' )
;

CREATE TABLE z_ServiceLedger ( payer_id int, clientvisit_id int, amount int, action_type varchar(50) ) ;
INSERT INTO z_ServiceLedger ( payer_id, clientvisit_id, amount, action_type )
VALUES (63,1,10,'XXXXX'),(63,2,20,'XXXXX'),(63,3,10,'XXXXX'),(63,4,30,'XXXXX') 
;

CREATE TABLE ClientVisit ( clientvisit_id int, client_id int, non_billable bit, duration int, emp_id int , rev_timein date, splitprimary_clientvisit_id int, gcode_primary_clientvisit_id int ) ;
INSERT INTO ClientVisit ( clientvisit_id, client_id, non_billable, duration, emp_id, rev_timein, splitprimary_clientvisit_id, gcode_primary_clientvisit_id )
VALUES
    (1, 1234, 0, 110, 1, getDate(), null, null ) 
  , (2, 1234, null, 120, 1, getDate(), null, null ) 
  , (3, 1234, 1, 110, 2, getDate(), null, null ) 
  , (4, 1234, 0, 130, 2, getDate(), null, null ) 
;

; WITH creds AS (
  SELECT c.credential_id, c.credentials, r.credential_rank
  FROM Credentials c 
  LEFT OUTER JOIN (VALUES (1,3),(2,2),(3,1) ) r(credential_id, credential_rank)
    ON c.credential_id = r.credential_id
)    
SELECT DISTINCT
    SUM(CASE WHEN ISNULL(v.non_billable,1) = 0 THEN v.duration ELSE 0 END)*1.0 / 60 AS billable_hours,
    SUM(CASE WHEN (ISNULL(v.non_billable,1) = 0 AND zsl.payer_id = 63) THEN v.duration ELSE 0 END)*1.0 / 60 AS billable_mro_hours,
    s2.credentials
FROM Z_ServiceLedger zsl
INNER JOIN ClientVisit v ON zsl.clientvisit_id = v.clientvisit_id
    AND v.rev_timein >= @param1
    AND v.rev_timein < DateAdd(d, 1, @param2) 
    AND v.splitprimary_clientvisit_id IS NULL
    AND v.gcode_primary_clientvisit_id IS NULL 
    AND ISNULL(v.non_billable,1) = 0
    AND v.duration*1.0 / 60 > 0 
    AND v.client_id <> 331771
INNER JOIN (
  SELECT s1.emp_id, s1.emp_name, s1.credential_id, s1.credentials, s1.endDate
  FROM (
    SELECT e.emp_id, e.emp_name, c.credential_id, c.credentials, ISNULL(ec.end_date,GETDATE()) AS endDate 
      , ROW_NUMBER() OVER (PARTITION BY e.emp_id ORDER BY c.credential_rank) AS rn
    FROM Employees e
    LEFT OUTER JOIN  EmployeeCredential ec ON e.emp_id = ec.emp_id
      AND ISNULL(ec.is_primary,1) <> 0 /* I don't think a NULL is_primary should be TRUE */
    LEFT OUTER JOIN creds c ON ec.credential_id = c.credential_id
  ) s1
  WHERE s1.rn = 1
) s2 ON v.emp_id = s2.emp_id
  AND v.rev_timein <= s2.endDate /* Credential not expired at rev_timein */
WHERE zsl.amount > 0 
  AND zsl.action_type NOT IN ('SERVICE RATE CHANGE', 'CLIENT STATEMENT') 
GROUP BY s2.credentials
ORDER BY s2.credentials

| billable_hours | billable_mro_hours | credentials |
|----------------|--------------------|-------------|
|       1.833333 |           1.833333 |          MA |
|       2.166666 |           2.166666 |         PhD |

主查询：
CREATE TABLE Employees ( emp_id int, emp_name varchar(20) ) ;
INSERT INTO Employees (emp_id, emp_name)
VALUES (1,'Jay'),(2,'Bob')
;

CREATE TABLE Credentials ( credential_id int, credentials varchar(20), credential_rank int ) ;
INSERT INTO Credentials (credential_id, credentials, credential_rank)
VALUES (1,'BA',3),(2,'MA',2),(3,'PhD',1)
;

CREATE TABLE EmployeeCredential (emp_id int, credential_id int, is_primary bit, end_date date )
INSERT INTO EmployeeCredential (emp_id, credential_id, is_primary, end_date)
VALUES 
    ( 1,2,null,'20200101' )
  , ( 1,3,0,'20200101' ) /* NON-PRIMARY */
  , ( 1,1,1,'20100101' ) /* EXPIRED CRED */
  , ( 2,3,null,'20200101' )
  , ( 2,3,1,'20200101' )
;

CREATE TABLE z_ServiceLedger ( payer_id int, clientvisit_id int, amount int, action_type varchar(50) ) ;
INSERT INTO z_ServiceLedger ( payer_id, clientvisit_id, amount, action_type )
VALUES (63,1,10,'XXXXX'),(63,2,20,'XXXXX'),(63,3,10,'XXXXX'),(63,4,30,'XXXXX') 
;

CREATE TABLE ClientVisit ( clientvisit_id int, client_id int, non_billable bit, duration int, emp_id int , rev_timein date, splitprimary_clientvisit_id int, gcode_primary_clientvisit_id int ) ;
INSERT INTO ClientVisit ( clientvisit_id, client_id, non_billable, duration, emp_id, rev_timein, splitprimary_clientvisit_id, gcode_primary_clientvisit_id )
VALUES
    (1, 1234, 0, 110, 1, getDate(), null, null ) 
  , (2, 1234, null, 120, 1, getDate(), null, null ) 
  , (3, 1234, 1, 110, 2, getDate(), null, null ) 
  , (4, 1234, 0, 130, 2, getDate(), null, null ) 
;

; WITH creds AS (
  SELECT c.credential_id, c.credentials, r.credential_rank
  FROM Credentials c 
  LEFT OUTER JOIN (VALUES (1,3),(2,2),(3,1) ) r(credential_id, credential_rank)
    ON c.credential_id = r.credential_id
)    
SELECT DISTINCT
    SUM(CASE WHEN ISNULL(v.non_billable,1) = 0 THEN v.duration ELSE 0 END)*1.0 / 60 AS billable_hours,
    SUM(CASE WHEN (ISNULL(v.non_billable,1) = 0 AND zsl.payer_id = 63) THEN v.duration ELSE 0 END)*1.0 / 60 AS billable_mro_hours,
    s2.credentials
FROM Z_ServiceLedger zsl
INNER JOIN ClientVisit v ON zsl.clientvisit_id = v.clientvisit_id
    AND v.rev_timein >= @param1
    AND v.rev_timein < DateAdd(d, 1, @param2) 
    AND v.splitprimary_clientvisit_id IS NULL
    AND v.gcode_primary_clientvisit_id IS NULL 
    AND ISNULL(v.non_billable,1) = 0
    AND v.duration*1.0 / 60 > 0 
    AND v.client_id <> 331771
INNER JOIN (
  SELECT s1.emp_id, s1.emp_name, s1.credential_id, s1.credentials, s1.endDate
  FROM (
    SELECT e.emp_id, e.emp_name, c.credential_id, c.credentials, ISNULL(ec.end_date,GETDATE()) AS endDate 
      , ROW_NUMBER() OVER (PARTITION BY e.emp_id ORDER BY c.credential_rank) AS rn
    FROM Employees e
    LEFT OUTER JOIN  EmployeeCredential ec ON e.emp_id = ec.emp_id
      AND ISNULL(ec.is_primary,1) <> 0 /* I don't think a NULL is_primary should be TRUE */
    LEFT OUTER JOIN creds c ON ec.credential_id = c.credential_id
  ) s1
  WHERE s1.rn = 1
) s2 ON v.emp_id = s2.emp_id
  AND v.rev_timein <= s2.endDate /* Credential not expired at rev_timein */
WHERE zsl.amount > 0 
  AND zsl.action_type NOT IN ('SERVICE RATE CHANGE', 'CLIENT STATEMENT') 
GROUP BY s2.credentials
ORDER BY s2.credentials

| billable_hours | billable_mro_hours | credentials |
|----------------|--------------------|-------------|
|       1.833333 |           1.833333 |          MA |
|       2.166666 |           2.166666 |         PhD |

有几件事需要注意：
1） 整数除法：duration/60
将返回一个整数。因此，如果您的持续时间为70，那么您的持续时间为70/60=1。您将错过10分钟，因为结果将转换回整数。你失去了额外的10分钟。可能不是你所说的。最简单的解决方案是将duration
乘以1.0
，这样它就被强制为十进制数据类型，而不会导致操作被视为整数
2） EmployeeCredential.is_primary！='False'
：不应考虑“True”/“False”字符串，而应使用实际的布尔值（1/0）。NULL
值应表示该值为非真
或假
，而不是表示真
。另外，在SQL中，=
将用于指示不等于
，但您应该改用
。它的意思是相同的，但在语法上更适合SQL
3） v.non\u billable=0和v.non\u billable='FALSE'
：这可以缩短为ISNULL（v.non\u billable，1）=0
，以使两个检查短路，特别是因为non\u billable
可以是NULL
。在比较数字0和字符串“False”时，还可以避免隐式类型转换
4） v.client\u id！='331771'
：更改为v.client_id33171
。首先是=到我前面提到的
。然后'331771'
被隐式转换为一个数字。您应该避免隐式转换
5） 您的组中最初有v.non_bilable
。由于未将其包括在选择中，因此不能使用它来按分组。此外，您已经过滤掉了除非计费=0
之外的所有内容，因此您永远不会有多个值